Let $\vec{A} = (1,2,3), \vec{B} = (3,-1,4), \vec{C}= (0,3,0) $
Calculate $\vec{A}$ $\cdot$ $(\vec{B} \times\vec{C})$ = $[\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}]$
I first calculate $\vec{A}$ $\cdot$ $(\vec{B} \times\vec{C})$ = $1 \times\begin{bmatrix}-1 & 4\\3 & 0 \end{bmatrix}$-2 $\begin{bmatrix}3 & 4\\0 & 0\end{bmatrix}$ + 3 $\begin{bmatrix}3 & -1\\0 & 3\end{bmatrix}$
$=1(-12)-2(0) +3(9)$ = $15$
I then try to calculate $\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}$, I know that $\vec{A}$ $\cdot$ $\vec{B}$ can be calculated as $a_xb_x+a_yb_y+a_zb_z$ I then try to similarly calculate $\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}$ in the same way as $a_xb_xc_x+a_yb_yc_y+a_zb_zc_z$ = $(1\times3 \times0) + (2 \times -1 \times 3) + (3 \times 4 \times 0) = 0 +-6+0 = -6$
$-6\neq 15$,
but $\vec{A}$ $\cdot$ $(\vec{B} \times \vec{C})$ = $\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}$
So I'm assuming that I'm calculating the dot product wrong... Any solutions hints or tips??
Here: The equality you are trying to prove is just a definition. The point is that there is no triple dot product as you intended in your question.